The magnitude of electric force exerted by a proton on an electron in a hydrogen atom is (Given distance of separation between charges, 1 Å)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.3 \times 10^{-8}\,\mathrm{N} \)
The force between two point charges is given by Coulomb's law. The proton and the electron carry charges of the same magnitude \( e \), so
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{r^{2}} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( r = 1\,\mathring{\mathrm{A}} = 10^{-10}\,\mathrm{m} \)
\( 1/4\pi\varepsilon_0 = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \)
\( \displaystyle F = \frac{9\times10^{9} \times (1.6\times10^{-19})^{2}}{(10^{-10})^{2}} \)
\( \displaystyle F = \frac{9\times10^{9} \times 2.56\times10^{-38}}{10^{-20}} = 2.3\times10^{-8}\,\mathrm{N} \)
The force is attractive, and its magnitude is \( 2.3\times10^{-8}\,\mathrm{N} \).